National Math Olympiad in Slovenia

Let $P$ be the intersection of the lines $MT$ and $KL$.
By Menelaus' theorem for the triangle $ALK$ and the colinear points $C, P$ and $M$ we have
$$\frac{AC}{CL} \cdot \frac{LP}{PK} \cdot \frac{KM}{MA} = -1.$$
In order to show that $|PL| = |PK|$ it suffices to see that
$$\frac{|AC|}{|CL|} = \frac{|MA|}{|MK|}. \qquad (12)$$
(Note: We will be using directed angles from this point on.)
The points $B, C, A$ and $M$ are concyclic, so $\nparallel BCL = -\nparallel ACB = \nparallel BMK$. The points $B, L, K$ and $A$ are also concyclic, so $\nparallel CLB = \nparallel ALB = \nparallel MKB$. The triangles $BCL$ and $BMK$ have two angles in common, so they are similar. It follows that
$$\frac{|CL|}{|MK|} = \frac{|BC|}{|BM|}. \qquad (13)$$
By the tangent-chord angle theorem we have $\nparallel TBC = \nparallel BMC$. The triangles $TBC$ and $TMB$ have two angles in common and are similar. Using the tangent-chord angle theorem again we see that $\nparallel CAT = \nparallel CMA$. So, the triangles $TAC$ and $TMA$ have two angles in common and are similar. From here we find the ratios $\frac{|BC|}{|BM|} = \frac{|TB|}{|TM|}$ and $\frac{|AC|}{|AM|} = \frac{|TA|}{|TM|}$. Combining the two with the equality between the tangent segments $|TA| = |TB|$ we get
$$\frac{|BC|}{|BM|} = \frac{|AC|}{|AM|}. \qquad (14)$$

(Note: The property (14) or $|AM| \cdot |BC| = |AC| \cdot |BM|$ is called the harmonic property of the cyclic quadrilateral.)
If we now combine the equalities (13) and (14), we get (12), which was our goal. From here we conclude that $P$ is the midpoint of $KL$ and so the midpoint of $KL$ does indeed lie on $MC$.